(a) Explain the terms reactance and impedance in an a.c circuit.
(b) A source of e.m.f. 240 v and frequency 50 Hz is connected to a resistor, an inductor and a capacitor in series. When the current in the capacitor is 10A, the potential difference across the resistor is 140V and that across the inductor is 50V. Draw the vector diagram of the potential difference across the inductor, the capacitor and the resistor.
Calculate the (i) potential difference across the capacitor; (ii) capacitance of the capacitor; (iii) inductance of the inductor.
(a) Reactance is the opposition in ohms to th alternating current by a capacitor or inductor. It is defined by
\(X_{C} = \frac{V}{I}\) or \(X_{L} = \frac{V}{I}\)
\(X_{C} = \frac{1}{2 \pi f c}\) or \(X_{L} = 2 \pi fl\)
V - voltage (V), c - current, f - frequency (Hz), I - inductance (H), c - capacitance (F).
Impedance is the opposition in ohms to the alternating current by combination of resistor, inductor or capacitor. It is defined as:
\(Z = \sqrt{R^{2} + X_{L} ^{2}}\)
\(Z = \sqrt{R^{2} + X^{2}}\)
\(Z = \sqrt{R^{2} + (X_{C} - X_{L})^{2}}\)
(b) 
Hence using 
,
\(V = \sqrt{V_{R} ^{2} + (V_{L} - V_{C})^{2}}\)
\(240 = \sqrt{140^{2} + (50 - V_{C})^{2}}\)
Square both sides,
\(57600 = 19600 + (50 - V_{C})^{2} \implies 57600 - 19600 = (50 - V_{C})^{2}\)
\(38000 = (50 - V_{C})^{2}\)
\(\sqrt{38000} = 50 - V_{C}\)
\(\pm 194.936 = 50 - V_{C} \implies V_{C} = \pm 194.936 + 50\)
\(V_{C} = -144.936 ; V_{C} = 244.936v\)
The voltage cannot be negative, hence, \(V_{C} = 244.936v\)
(ii) Using \(X_{C} = \frac{1}{2 \pi fc} = \frac{V}{I}\)
\(\frac{244.936}{10} = \frac{1}{2\pi \times 50 \times c}\)
\(24.4936 = \frac{1}{2\pi \times 50 \times c} \implies c = \frac{1}{2\pi \times 50 \times 24.4936}\)
\(c = \frac{1}{7694.89} = 0.00013F = 130\mu F\)
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